Question

A 645-turn coil with a 20.250 m ​2 ​​ area is spun in Earth’s 5.00×10 ​−5 ​​ T magnetic field, producing a 1.25-V maximum emf. At what angular velocity must the coil be spun?

Answer 1

Answer:

$\omega = 1.914\ rad/s$

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

$\varepsilon = NAB\omega$

$\omega= \dfrac{\varepsilon}{NAB}$

$\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}$

$\omega = 1.914\ rad/s$

Angular velocity of the coil = $\omega = 1.914\ rad/s$